What is the integral of (2x)/((2x+1)) ?

The integral of (2x)/((2x+1)) is x+1/2-1/2 ln|2x+1|+C

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integral of (2x)/((2x+1))

integral

\frac{2 x}{( 2 x + 1 )}

x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣ + C

Solution steps

\int \frac{2 x}{2 x + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{x}{2 x + 1} d x

Apply u-substitution

: \int \frac{u - 1}{4 u} d u

= 2 \cdot \int \frac{u - 1}{4 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{4} \cdot \int \frac{u - 1}{u} d u

Expand

\frac{u - 1}{u} : 1 - \frac{1}{u}

= 2 \cdot \frac{1}{4} \cdot \int 1 - \frac{1}{u} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 2 \cdot \frac{1}{4} (\int 1 d u - \int \frac{1}{u} d u)

\int 1 d u = u

\int \frac{1}{u} d u = ln ∣ u ∣

= 2 \cdot \frac{1}{4} (u - ln ∣ u ∣)

Substitute back

u = 2 x + 1

= 2 \cdot \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣)

Simplify

2 \cdot \frac{1}{4} (2 x + 1 - ln ∣ 2 x + 1 ∣) : x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣

= x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= x + \frac{1}{2} - \frac{1}{2} ln ∣ 2 x + 1 ∣ + C

\frac{1}{x ^{2}} lo g_{10} (x)

2 t + 3

\frac{( x ^{2} + 1 ) e ^{x}}{( x + 1 ) ^{2}}

e^{6 x}

\frac{x}{( 2 + 5 x )}

What is the integral of (2x)/((2x+1)) ?
The integral of (2x)/((2x+1)) is x+1/2-1/2 ln|2x+1|+C